[Stoves] Hand Crank Dynamo for Fanned Stove?

[Charlie Sellers]

These are my general finds as well, esp. if you combine this email with David's about super bright LEDs (that they are ~1 W, so the crank+battery combo is somewhat meant for a load this size) - DC-DC is always possible to reach other voltages, and there are 3 batteries in series for 3.6 volts.  Measuring the state of charge of a battery is a little finicky so I have not bothered with "cranks per second of output to fan"), tracking down other things instead - mine has an AC charger, takes all night to charge (3600 mA-hr capacity, but we can't access all of it) it seems, and with full charge it will provide several hours of fan power at full speed (and will take larger loads for shorter amounts of time).  How many cranks it takes to fully charge... I don't even want to think of it but the manufacturer may know.

Good analysis mate, and I'll have more too.

Charlie


Robert Penn Taylor <rptaylor at iastate.edu> wrote: Paul, David, Crispin, and the other half-dozen folks who've weighed in 
on this topic:

David G. LeVine wrote:
> At 06:49 PM 12/4/2007, you wrote:
> I just looked at three "muffin" fans (computer fans.)  They were "5V, 
> 0.55W" or 0.11 Amperes, "12 Volt, 0.08 Amperes" and "12 Volt, 0.12 
> Amperes"  Now, a typical LED runs about 3.5 Volts for white and can 
> be in the 20 mA to 1,500 mA range.  The 3 LED flashlight is PROBABLY 
> running either no more than 10.5 Volts OR has a complex 
> driver,  

More likely the LEDs are wired in parallel, and the battery provides 3.5 
Volts.

After typing that last sentence, I dug out my 3-LED handcrank flashlight 
and took it apart. Here's the story: the LEDs are wired in parallel. The 
battery is 3 button-style NiMHs wired in series, the total stack 
nominally providing 3.6V. The dynamo provides ~5V (in case it isn't 
obvious, the battery "charger" -- the dynamo in this case -- must 
provide a higher voltage than the open voltage of the battery or no 
charging takes place.)

Now, the batteries in this flashlight are rated at 80mAh capacity, so 
3.6V * 80mAh * 3600s/h = 1037 Joules. When you're looking at battery 
_capacity_, that's the important metric: Joules. Not amp-hours, not 
volts, not amps, not Watts, but Joules. This tells us how much energy 
can be stored by the battery. It comes out at some set voltage (in 
theory, but reality is a little different), but we can easily transform 
that voltage to a higher or lower voltage, so it's really Joules that we 
care about.

Paul said his fan requires 4 Watts. The question now becomes how long 
will it takes to dissipate 1037 Joules at 4 watts (Joules/second)? Well, 
that's 1037/4 = 260 second = ~4.3 minutes. That's if the battery in this 
flashlight is fully charged. The other question about whether 3.6V will 
run the fan is less important because we can always transform the 
voltage with a DC-DC transformer (which, it's true, will eat away some 
of the energy coming through.) The 0.55W muffin fan David mentions would 
give 1037J/0.55W = ~30 minutes.




       
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